### Video Transcript

Find the points on the curve π¦ squared is equal to two π₯ plus 21 that are closest to the point negative six, zero.

The question gives us the equation of a curve. And it wants us to find the points that are on this curve which are closest to the point negative six, zero. To start, letβs recall the formula for the distance between a point π₯, π¦ and a point π, π. By the Pythagorean theorem, we have that this distance is equal to the square root of π₯ minus π squared plus π¦ minus π squared.

In our case, the point π₯, π¦ lies on our curve and the point π, π can be negative six, zero. So, using this formula, if π₯, π¦ lies on our curve and π, π is negative six, zero, then the length of the line between the point π₯, π¦ and the point negative six, zero is equal to the square root of π₯ minus negative six squared plus π¦ minus zero squared.

And we could simplify this expression. Distributing both of the squares over our parentheses, we get π₯ squared plus 12π₯ plus 36 and π¦ squared. But remember, the point π₯, π¦ lies on our curve, so π¦ squared is actually equal to two π₯ plus 21. So, if we substitute π¦ squared is equal to two π₯ plus 21, we get the length is equal to the square root of π₯ squared plus 12π₯ plus 36 plus two π₯ plus 21. And of course, we can simplify this expression further.

12π₯ plus two π₯ is equal to 14π₯. And 36 plus 21 is equal to 57. So, we found a formula for the length of the line between the point π₯, π¦ on our curve and the point negative six, zero. Itβs equal to the square root of π₯ squared plus 14π₯ plus 57. This is a function entirely in terms of π₯. And the question wants this length to be as small as possible. This is an optimization problem, so weβll solve this by finding the critical points of our length function.

We see that our formula for the length is the composition of two functions; itβs the square root of a polynomial. So, we want to differentiate this by using the chain rule. The chain rule tells us if π’ is a function of π£ and π£ in turn is a function of π₯, then the derivative of π’ of π£ of π₯ is equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯. So, weβll set π£ of π₯ to be our inner function. Thatβs the polynomial π₯ squared plus 14π₯ plus 57.

So, this gives us the length of our line π in terms of π₯ is equal to the square root of π£ of π₯. Next, weβll set π’ of π£ to be the square root of π£. So, doing this, weβve rewritten our length to be π’ composed with π£. π’ is a function of π£, and π£ in turn is a function of π₯. So, to use the chain rule, we need expressions for π’ prime and π£ prime. Letβs start with π’ prime.

Itβs the derivative of the square root of π£ with respect to π£. Using our laws of exponents, we can rewrite the root of π£ as π£ to the power of one-half. And we can differentiate this by using the power rule for differentiation. We multiply by the exponent of π£, which is one-half, and reduce this exponent by one. This gives us one-half π£ to the power of negative one-half, which weβll rewrite as one over two root π£.

Next, we want to find an expression for π£ prime of π₯. Thatβs the derivative of π₯ squared plus 14π₯ plus 57 with respect to π₯. Weβll do this by using the power rule for differentiation on each term. This gives us two π₯ plus 14. Now, by using the chain rule, we have π prime of π₯ is equal to π£ prime of π₯ times π’ prime of π£ of π₯. Substituting in our expressions for π£ prime, π’ prime, and π£ of π₯, we get that π prime of π₯ is equal to two π₯ plus 14 divided by two times the square root of π₯ squared plus 14π₯ plus 57.

And we can simplify this slightly. Weβll cancel the shared factor of two in our numerator and our denominator. This gives us that π prime of π₯ is equal to π₯ plus seven divided by the square root of π₯ squared plus 14π₯ plus 57. Remember, we want to find the critical points of this function. Thatβs where the derivative is equal to zero or where the derivative does not exist. Letβs start by checking for points where the derivative will not exist.

The derivative will not exist if the denominator is equal to zero or if weβre taking the square root of a negative number. To check if this happens, weβll complete the square for our quadratic. To complete the square, we want π₯ plus half the coefficient of our π₯ term all squared. This means when we expand the parentheses on our right-hand side, weβll be adding an extra 49. So, we need to subtract this 49. So, currently expanding and simplifying our right-hand side, we get π₯ squared plus 14π₯.

So, to balance the two sides of this equation, we just need to add 57. And negative 49 plus 57 is equal to eight. So, by completing the square, weβve rewritten the quadratic in our denominator as π₯ plus seven all squared plus eight. Letβs write this into our derivative function. So, weβve now rewritten π prime of π₯ to be π₯ plus seven divided by the square root of π₯ plus seven squared plus eight. And the reason we did this is it makes finding our critical points easier.

π₯ plus seven all squared is a square, so itβs greater than or equal to zero. But then we add eight to this value. So, our denominator is always positive. So, weβre always taking the square root of a positive number. Therefore, our denominator is always positive. So, our derivative is defined for all values of π₯. So, the only critical points will be where our derivative is equal to zero. This means our numerator must be equal to zero, which means that π₯ is equal to negative seven. So, weβve shown that our function has one critical point where π₯ is equal to negative seven.

We need to decide what type of critical point this is. We could do this by using the first or second derivative test. However, in this case, weβll do this by looking at our function π prime of π₯. Weβve already explained that the denominator of π prime of π₯ is positive for all values of π₯. So, whether π prime of π₯ is positive or negative depends entirely on the sign of our numerator. And our numerator is negative if π₯ is less than negative seven. And our numerator is positive if π₯ is greater than negative seven.

So, letβs think what this means graphically for a second. Our slope at negative seven is zero. For all values of π₯ less than negative seven, our slope is defined and itβs negative. And for all values of π₯ greater than seven, our slope is defined and positive. So, we have a continuous function with a continuous derivative with one turning point at π₯ is equal to negative seven. So, in actual fact, this is a global minimum of our function.

The last thing to do is to find the coordinates of any points where π₯ is equal to negative seven. Weβll do this by substituting π₯ is equal to negative seven into our equation for the curve. Substituting π₯ is equal to negative seven into the equation of our curve, we get π¦ squared is equal to two times negative seven plus 21, which simplifies to give us π¦ squared is equal to seven. And then, we just take the square root of both sides of this equation, remembering we get a positive and a negative square root. We get π¦ is equal to positive or negative root seven. Therefore, weβve shown the points on the curve π¦ squared is equal to two π₯ plus 21 that are closest to the point negative six, zero are the points with coordinates negative seven, root seven and negative seven, negative root seven.